Related rates — setting up and solving: Question bank

4.1.25 · D5 · Maths › Calculus I — Limits & Derivatives › Related rates — setting up and solving

Har item ek Question ::: Answer reveal hai. Answer cover karo, decide karo, phir check karo.

Traps ko concrete rakhne ke liye, yahan woh do geometries hain jis par items baar baar wapas aate hain, draw karke aur label karke jaise Step 1 of the recipe demand karta hai. Jab bhi koi item $x, y$ ya $h, r$ ka naam le, inhe refer karo.

Figure — Related rates — setting up and solving

True or false — justify

Agar tum differentiate karne se pehle instant ke numbers plug in kar do, toh bhi sahi rate milega jab tak baad mein carefully differentiate karo.

False. Ek baar jab ek changing variable fixed number ban jaata hai, uska derivative

0

ho jaata hai, toh differentiating baad mein silently woh rate drop kar deta hai jise tum dhundh rahe the.

Ladder problem mein, ladder ka top constant speed se neeche jaata hai kyunki base constant speed se bahar jaata hai.

False.

\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t}

ratio

x / y

par depend karta hai, jo ladder ke khisakte waqt change hota rehta hai — toh even constant base speed ke saath top accelerate karta hai.

A = π r^{2}

ke liye, rate

\frac{d A}{d t}

har instant par same hoga agar

\frac{d r}{d t}

constant hai.

False.

\frac{d A}{d t} = 2 π r \frac{d r}{d t}

r

ke saath badhta hai: ek bada circle har second zyada rim add karta hai, toh area rate fixed radial speed par bhi badhti rehti hai.

Related rates aur implicit differentiation essentially same procedure hain.

True. Dono ek relation ko differentiate karte hain jab variables ko ek hidden parameter ke functions ki tarah treat kiya jaata hai — implicit differentiation

x

use karta hai, related rates time

t

use karta hai. Dekho Implicit differentiation.

Cone problem mein

\frac{d h}{d t}

ki ek negative value ka matlab hoga ki tank drain ho raha hai.

True. Ek rate ka sign uski direction hai; negative depth-rate ka matlab hai water level gir raha hai, toh us instant par volume decrease ho raha hai.

Related rates mein chain rule optional hai — tum hamesha ek variable ko

t

ke terms mein solve karke directly differentiate kar sakte ho.

False in practice. Tumhare paas almost kabhi koi variable

t

ke explicit function ke roop mein nahi hota; Chain rule precisely woh tool hai jo tum bina us formula ke differentiate karne deta hai.

Agar do quantities ek equation se linked hain lekin us equation mein explicitly koi time-dependent nahi hai, toh koi bhi time ke saath change nahi karta.

False. Equation unki values constrain karta hai, unke time-behaviour ko nahi; dono

t

ke functions ho sakte hain — woh hidden time-dependence related rates ka poora engine hai.

Spot the error

"Oil circle mein

r = 10

hai, toh

A = π (100) = 100 π

; isliye

\frac{d A}{d t} = 100 π

Student ne ek number differentiate kiya, rate nahi.

100 π

area ki value hai; rate ke liye

\frac{d A}{d t} = 2 π r \frac{d r}{d t}

chahiye, jo diya gaya

\frac{d r}{d t}

use karta hai, sirf

r

nahi.

"Ladder:

x^{2} + y^{2} = 25

, differentiate karke

2 x + 2 y = 0

milta hai."

Wo Chain rule se

\frac{d x}{d t}

aur

\frac{d y}{d t}

factors bhool gaye. Sahi:

2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0

"Cone:

V = \frac{1}{3} π r^{2} h

, differentiate karke

\frac{d V}{d t} = \frac{1}{3} π r^{2} \frac{d h}{d t}

Unhone

r

ko constant treat kiya.

r

aur

h

dono change karte hain, toh ya toh Similar triangles ke zariye pehle

r

eliminate karo, ya

r^{2} h

par product rule apply karo.

"Ladder ke top ka rate negative aaya, toh maine use positive kar diya kyunki speed negative nahi ho sakti."

Sign drop karna physics ko destroy kar deta hai. Negative sign information hai: yeh kehta hai top neeche ja raha hai. Speed

∣ \frac{d y}{d t} ∣

hai, lekin signed rate apna sign rakhna chahiye.

\frac{d}{d t} (x^{2}) = 2 x

Inner rate missing hai. Kyunki

x = x (t)

hai,

\frac{d}{d t} (x^{2}) = 2 x \frac{d x}{d t}

; sirf

2 x

likhna pretend karta hai ki hum

x

ke respect mein differentiate kar rahe hain,

t

ke nahi.

"Ek chalta hua insaan ki shadow ke liye:

x

ko us insaan ki lamppost se distance maano aur

s

ko shadow ki tip ki distance, toh roshni, head aur tip do similar triangles banate hain. Maine pehle

x = 5

m set kiya, phir similar-triangle relation likha aur differentiate kiya."

Distance

x = 5

ko pehle fix karna, relation banane aur differentiate karne se pehle, us variable ko freeze kar deta hai jiska rate

\frac{d x}{d t}

(walking speed) matter karta hai — wahi deadly early-substitution error, bas disguised.

\frac{d V}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t}

, aur kyunki

\frac{d V}{d t} = 2

constant hai,

\frac{d h}{d t}

bhi constant hai."

\frac{d h}{d t} = \frac{8}{π h ^{2}}

jaise

h

badhta hai shrink karta hai — ek constant inflow ek widening cone mein constant rise nahi deta.

Why questions

Hum linking equation ko

x

r

ke respect mein nahi balki

t

ke respect mein differentiate kyun karte hain?

Kyunki hum time rates (

\frac{d}{d t}

) chahte hain. Time woh shared parameter hai jis par har quantity depend karti hai, toh

t

mein differentiate karna un actual speeds ke beech relations produce karta hai jinke baare mein puchha ja raha hai.

Cone problem mein differentiate karne se pehle

r

eliminate kyun karna chahiye?

Dono

r

aur

h

rakhne se equation mein ek unknown

\frac{d r}{d t}

aa jaata hai. Similar triangles use karke pehle

r = \frac{h}{2}

likhna sirf woh rate chodta hai jise hum solve kar sakte hain.

Sliding-ladder mein top speed infinity ki taraf kyun blow up karti hai jab ladder flat hone wali hoti hai?

\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t}

mein, height

y \to 0

denominator mein hoti hai, toh rate

\to - \infty

— ek limiting behaviour, real infinite speed nahi (model pehle break ho jaata hai).

Kisi bhi algebra se pehle, ek labelled picture draw karna Step 1 kyun hai?

Picture reveal karti hai ki variables ko kaunsa geometric law link karta hai (Pythagoras, similar triangles, volume formula) — tum ek linking equation choose nahi kar sakte jo tumne dekhi nahi.

Parent ke recipe mein product rule nahi balki chain rule ko top billing kyun milti hai?

Har single variable composite hai — ek quantity of time ki quantity — toh Chain rule har term par fire karta hai; product rule sirf tab aata hai jab do changing variables multiply karte hain.

Related rates ka jawab kisi instant par exactly

0

kyun ho sakta hai jab sab kuch move kar raha ho?

Ek rate

0

hota hai jab geometry us coordinate mein momentarily stationary hoti hai — jaise ek arc ke top par point ka vertical rate zero hota hai jabki horizontally move karta rehta hai.

Edge cases

\frac{d A}{d t} = 2 π r \frac{d r}{d t}

bilkul start mein kya hoga, jab

r = 0

ho?

Yeh

0

equals karta hai: ek point-sized spill mein push out karne ke liye koi rim nahi hoti, toh area rate momentarily zero hoti hai chahe radius already badh raha ho.

Ladder problem mein, jab base exactly wall par ho (

x = 0

) us instant par

\frac{d y}{d t}

kya hai?

\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} = 0

: jab ladder vertical ho toh top momentarily motionless hota hai girne se pehle.

Cone formula

\frac{d h}{d t} = \frac{8}{π h ^{2}}

depth-rate ke liye kya predict karta hai jab tank almost empty ho,

h \to 0^{+}

Rate

\to + \infty

— sharp apex par thoda sa paani ek bahut patli layer bahut jaldi bhar deta hai, toh level initially tezi se upar jaati hai.

Agar

\frac{d r}{d t} = 0

(radius momentarily frozen) ho lekin spill ki area ho, toh kya

\frac{d A}{d t} = 0

hai?

Haan:

\frac{d A}{d t} = 2 π r \cdot 0 = 0

. Zero radial speed ke instant par area momentarily change nahi ho rahi, chahe circle kitna bhi bada kyun na ho.

Kya cone mein

\frac{d V}{d t}

positive ho sakta hai jabki

\frac{d h}{d t}

negative ho?

Nahi — is cone ke liye

\frac{d V}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t}

hai aur

h > 0

ke liye

\frac{π h ^{2}}{4} > 0

hai, toh dono rates hamesha same sign share karte hain; woh sirf saath mein hi vanish ho sakte hain (

h \to 0

ki limit mein).

Agar linking equation mein ek constant involve ho (jaise fixed ladder length

L

)? Uska time-derivative kya hoga?

Kisi bhi constant ka derivative

0

hota hai:

\frac{d}{d t} (L^{2}) = 0

. Wahi vanishing ladder equation ke right side par "

= 0

" produce karta hai.

Recall Ek-line survival summary

Traps ::: variables ko differentiate karne tak alive rakho, sign kabhi drop mat karo, aur linking equation supply karne ke liye hamesha geometry ko (number ko nahi) pehle aane do.

Connections

Related rates — setting up and solving — woh parent jise yeh bank stress-test karta hai.
Chain rule — woh engine jis par yahan har trap circle back karta hai.
Implicit differentiation — disguise mein twin technique.
Similar triangles — cone traps ke peechhe elimination trick.
Pythagorean theorem — ladder ka linking law.
Derivatives as rates of change — kyun signs directions mean karte hain.
Optimization — jahan stationary rates ( $= 0$ ) naye meaning ke saath wapas aate hain.